Question

prove ...................​

Answer 1

$\frac{ \tan \alpha + \sin \alpha }{ \tan\alpha - \sin \alpha } = \frac{ \sec \alpha + 1 }{ \sec \alpha - 1 }$

On taking L.H.S. :

$\frac{ \tan\alpha + \sin\alpha }{ \tan \alpha - \sin \alpha } \\ \\ = > \frac{ \frac{ \sin \alpha }{ \cos \alpha } + \sin \alpha }{ \frac{ \sin \alpha }{ \cos \alpha } - \sin \alpha } \\ \\ = > \frac{ \sin \alpha ( \frac{1}{ \cos \alpha } + 1)}{ \sin \alpha ( \frac{1}{ \cos \alpha } - 1) } \\ \\ = > \frac{ \sec \alpha + 1 }{ \sec \alpha - 1 }$

=> R.H.S.

HENCE PROVED

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Answer 2

Answer:

Step-by-step explanation:

mark me i needed

The answer is here,

Given that,

= > \: \frac{a + ib}{c + id} = p + iq

We can replace the " i " as " -i ".

= > \frac{a - ib}{c - id} = p - iq